\(=>x=25:\left(\dfrac{1}{2}-\dfrac{1}{5}\right)=25:\dfrac{3}{10}=\dfrac{250}{3}\)

\(\dfrac{x}{2}-\dfrac{x}{5}=25\)

\(\dfrac{5x}{10}-\dfrac{2x}{10}=25\)

\(\dfrac{5x-2x}{10}=25\)

\(\dfrac{3x}{10}=25\)

\(3x=10.25\)

\(3x=250:3\)

\(3x=\dfrac{250}{3}\)

j: (2x-1)(3x+1)-(4-3x)(3-2x)=3

=>(2x-1)(3x+1)-(3x-4)(2x-3)=3

=>\(6x^2+2x-3x-1-\left(6x^2-9x-8x+12\right)=3\)

=>\(6x^2-x-1-\left(6x^2-17x+12\right)=3\)

=>\(6x^2-x-1-6x^2+17x-12=3\)

=>16x-13=3

=>16x=16

=>x=1

k: \(\left(2x+1\right)\left(x+3\right)-\left(x-5\right)\left(2x+7\right)=8\)

=>\(2x^2+6x+x+3-\left(2x^2+7x-10x-35\right)=8\)

=>\(2x^2+7x+3-2x^2-7x+10x+35=8\)

=>10x+38=8

=>10x=-30

=>x=-3

m: 2(3x-1)(2x+5)-6(2x-1)(x+2)=-6

=>\(2\left(6x^2+15x-2x-5\right)-6\left(2x^2+4x-x-2\right)=-6\)

=>\(2\left(6x^2+13x-5\right)-6\left(2x^2+3x-2\right)=-6\)

=>\(12x^2+26x-10-12x^2-18x+12=-6\)

=>8x+2=-6

=>8x=-8

=>x=-1

1/ (2x+3)(x-4)+(x+5)(x-2)=(3x-5)(x-4)

<=> 2x² - 8x + 3x - 12 + x² - 2x + 5x  - 10 - 3x² + 12x + 5x - 20 = 0

<=> 15x - 20 = 0

<=> 15x = 20

<=> x = 4/3

1: \(\left(x-3\right)\left(2x-5\right)-3x\left(x+4\right)\)

\(=2x^2-5x-6x+15-3x^2-12x\)

\(=-x^2-23x+15\)

2: \(\left(\dfrac{1}{2}x+5\right)\left(2x-\dfrac{1}{5}\right)\)

\(=x^2-\dfrac{1}{10}x+10x-1\)

\(=x^2+\dfrac{99}{10}x-1\)

\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)

\(\Leftrightarrow-3x=-\frac{13}{4}\)

\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)

\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)

\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)

\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)

\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)

\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)

\(\Leftrightarrow x=-\frac{6}{11}\)

d,e,f Tương tự

7: Ta có: \(\left(3x+4\right)\left(2x-1\right)+6x\left(1-x\right)=0\)

\(\Leftrightarrow6x^2-3x+8x-4+6x-6x^2=0\)

\(\Leftrightarrow11x=4\)

hay \(x=\dfrac{4}{11}\)

8: Ta có: \(2x\left(x^2-1\right)+x\left(-2x^2-3x+1\right)=-x-27\)

\(\Leftrightarrow2x^3-2x-2x^3-3x^2+x+x+27=0\)

\(\Leftrightarrow x^2=9\)

hay \(x\in\left\{3;-3\right\}\)